Question

If \(A\) and \(B\) simultaneously start walking East and South respectively, then which of the following is true of the distance of closest approach \(d_1\) between them?

• A. \(d_1>5\) m
• B. \(d_1 <5\) m
• C. \(d_1 = 5\) m
• D. Cannot be determined

Hint:

To find minimum distance between two moving points, model positions with time, square the distance expression and minimize the resulting quadratic.

Answer 1

The Correct Option is B

Step 1: Coordinates 
Point \(A = (5, 4)\), walking East \(\Rightarrow\) direction = \((1, 0)\), speed = 1.4 m/s 
Point \(B = (15, 24)\), walking South \(\Rightarrow\) direction = \((0, -1)\), speed = 2.1 m/s 

Step 2: Position equations 
At time \(t\), \[ A(t) = (5 + 1.4t, 4), \quad B(t) = (15, 24 - 2.1t) \] 

Step 3: Distance function 
\[ d^2(t) = [(5 + 1.4t) - 15]^2 + [4 - (24 - 2.1t)]^2 
= (-10 + 1.4t)^2 + (-20 + 2.1t)^2 \] \[ d^2(t) = (100 - 28t + 1.96t^2) + (400 - 84t + 4.41t^2) 
= 500 - 112t + 6.37t^2 \] 

Step 4: Minimize distance 
Minimize quadratic: \[ d^2(t) = 6.37t^2 - 112t + 500 \Rightarrow t = \frac{112}{2 \cdot 6.37} \approx 8.79 \] \[ d^2_{min} = 6.37(8.79)^2 - 112(8.79) + 500 \approx 24.3 \Rightarrow d_{min} \approx \sqrt{24.3} \approx 4.93 \] \[ \boxed{d_1 <5} \]